- #1
Theage
- 11
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Note: I'm posting this in the Quantum Physics forum since it doesn't really apply to HEP or particle physics (just scalar QFT). Hopefully this is the right forum.
In Peskin and Schroeder, one reaches the following equation for the spacetime Klein-Gordon field:
$$\phi(x,t)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\Big(a(p)e^{-ip\cdot x}+a^\dagger(p)e^{ip\cdot x}\Big)$$
Then they say that the propagation amplitude for a particle to go from a spacetime point x to a spacetime point y is [itex]\langle 0\vert\phi(x)\phi(y)\vert 0\rangle[/itex] where the ket |0> is the vacuum. I understand this up to here. But then they start computing it as follows (equation 2.50):
$$\langle 0\vert\phi(x)\phi(y)\vert 0\rangle = \int\frac{d^3p\,d^3q}{(2\pi)^6}\frac{1}{2\sqrt{E_{p}E_{p'}}}\Big(a(p)e^{-ipx}+a^\dagger(p)e^{ipx}\Big)\Big(a(q)e^{-iqy}+a^\dagger(q)e^{iqy}\Big)$$
Clearly there will now be four terms when you expand the parentheses, and the book claims that all of these vanish except for the term with [itex]\langle 0\vert a(p)a^\dagger(q)\vert 0\rangle[/itex]. Two questions:
a) Wouldn't this term vanish also since a(p) kills the vacuum bra, producing a zero?
b) Why doesn't the term with [itex]a^\dagger(p)a^\dagger(q)[/itex] stay? In that term there are no annihilation operators to kill the vacuum, so surely the term should vanish.
In Peskin and Schroeder, one reaches the following equation for the spacetime Klein-Gordon field:
$$\phi(x,t)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\Big(a(p)e^{-ip\cdot x}+a^\dagger(p)e^{ip\cdot x}\Big)$$
Then they say that the propagation amplitude for a particle to go from a spacetime point x to a spacetime point y is [itex]\langle 0\vert\phi(x)\phi(y)\vert 0\rangle[/itex] where the ket |0> is the vacuum. I understand this up to here. But then they start computing it as follows (equation 2.50):
$$\langle 0\vert\phi(x)\phi(y)\vert 0\rangle = \int\frac{d^3p\,d^3q}{(2\pi)^6}\frac{1}{2\sqrt{E_{p}E_{p'}}}\Big(a(p)e^{-ipx}+a^\dagger(p)e^{ipx}\Big)\Big(a(q)e^{-iqy}+a^\dagger(q)e^{iqy}\Big)$$
Clearly there will now be four terms when you expand the parentheses, and the book claims that all of these vanish except for the term with [itex]\langle 0\vert a(p)a^\dagger(q)\vert 0\rangle[/itex]. Two questions:
a) Wouldn't this term vanish also since a(p) kills the vacuum bra, producing a zero?
b) Why doesn't the term with [itex]a^\dagger(p)a^\dagger(q)[/itex] stay? In that term there are no annihilation operators to kill the vacuum, so surely the term should vanish.